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Q. 7

Expert-verified
Found in: Page 209

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Differentiate $f\left(x\right)={\left(3x+\sqrt{x}\right)}^{2}$in three ways. When you have completed all three parts, show that your three answers are the same: (a) with the chain rule (b) with the product rule but not the chain rule (c) without the chain or product rules.

The derivative of $f\left(x\right)$ using each rule is:localid="1648368040492" $f\text{'}\left(x\right)=18x+9\sqrt{x}+1$

See the step by step solution

## Step 1. Given information:

The function is:

$f\left(x\right)={\left(3x+\sqrt{x}\right)}^{2}$

## Part (a). Step 1. To find derivative using chain rule:

$f\left(x\right)={\left(3x+\sqrt{x}\right)}^{2}$

Let $u=3x+\sqrt{x}$

then $f\left(u\right)={u}^{2}$

Derivative of $u$ with respect to $x$:

$\frac{du}{dx}=\frac{d}{dx}\left(3x+\sqrt{x}\right)\phantom{\rule{0ex}{0ex}}=3+\frac{1}{2\sqrt{x}}$

Derivative of $f$ with respect to $u$:

$\frac{df}{du}=\frac{d}{du}\left({u}^{2}\right)\phantom{\rule{0ex}{0ex}}=2u\phantom{\rule{0ex}{0ex}}=2\left(3x+\sqrt{x}\right)$

Derivative of $f$ with respect to $x$:

localid="1648368008391" $\frac{df}{dx}=\frac{df}{du}×\frac{du}{dx}\phantom{\rule{0ex}{0ex}}=2\left(3x+\sqrt{x}\right)\left(3+\frac{1}{2\sqrt{x}}\right)\phantom{\rule{0ex}{0ex}}=2\left(9x+\frac{3}{2}\sqrt{x}+3\sqrt{x}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=2\left(9x+\frac{9}{2}\sqrt{x}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=18x+9\sqrt{x}+1\phantom{\rule{0ex}{0ex}}$

## Part (b). Step 1. To find derivative using product rule:

$f\left(x\right)={\left(3x+\sqrt{x}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(3x+\sqrt{x}\right).\left(3x+\sqrt{x}\right)$

$U=\left(3x+\sqrt{x}\right)\phantom{\rule{0ex}{0ex}}V=\left(3x+\sqrt{x}\right)$

localid="1648367311339" $\frac{dU}{dx}=\frac{d}{dx}\left(3x+\sqrt{x}\right)\phantom{\rule{0ex}{0ex}}=3+\frac{1}{2\sqrt{x}}\phantom{\rule{0ex}{0ex}}=\frac{dV}{dx}$

Now derivative of a function using the product rule is:

localid="1648367986645" $\frac{df}{dx}=U\frac{dV}{dx}+V\frac{dU}{dx}\phantom{\rule{0ex}{0ex}}=\left(3x+\sqrt{x}\right)\left(3+\frac{1}{2\sqrt{x}}\right)+\left(3x+\sqrt{x}\right)\left(3+\frac{1}{2\sqrt{x}}\right)\phantom{\rule{0ex}{0ex}}=2\left(3x+\sqrt{x}\right)\left(3+\frac{1}{2\sqrt{x}}\right)\phantom{\rule{0ex}{0ex}}=2\left(9x+\frac{3}{2}\sqrt{x}+3\sqrt{x}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=2\left(9x+\frac{9}{2}\sqrt{x}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=18x+9\sqrt{x}+1$

## Part (c). Step 1. To find derivative without using chain rule or product rule:

$f\left(x\right)={\left(3x+\sqrt{x}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(3x\right)}^{2}+{\left(\sqrt{x}\right)}^{2}+2\left(3x\right)\left(\sqrt{x}\right)\phantom{\rule{0ex}{0ex}}=9{x}^{2}+x+6{x}^{\frac{3}{2}}$

Now differentiate this with respect to $x$:

$\frac{df}{dx}=\frac{d}{dx}\left(9{x}^{2}+x+6{x}^{\frac{3}{2}}\right)\phantom{\rule{0ex}{0ex}}=9.2x+1+6.\frac{3}{2}{x}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=18x+1+9\sqrt{x}$